Hi, I'd appreciate some help with this question. I'm not sure where to start.
The first five terms of an arithmetic sequence are:
x+1, 2x, 2(2x+3)/6x, x^22, 5x3
Show that the term 4x^23 is not in the sequence.

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 02032018 18:03

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 02032018 18:14
it is barely conceivable that this is a GCSE question
smh
anyhow....
you can find a rule in the usual way;
the gap between the first two terms is x  1
so (x  1) *N + ?
letting N = 1 we can see that ? must be 2
so the n^{th} term is (x  1)*N + 2
see what happens if you let 4x^{2}  3 equal (x  1)*N + 2 
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 02032018 18:28
(Original post by the bear)
it is barely conceivable that this is a GCSE question
smh
anyhow....
you can find a rule in the usual way;
the gap between the first two terms is x  1
so (x  1) *N + ?
letting N = 1 we can see that ? must be 2
so the n^{th} term is (x  1)*N + 2
see what happens if you let 4x^{2}  3 equal (x  1)*N + 2
Just using a term in the sequence to show whether I know a term fits:
2x=(x1)n+2
2x2=(x1)n
2(x1)=(x1)n
So n=2, so fits the sequence.
In the case of 4x^23:
4x^23=(x1)n+2
4x^25=(x1)n
4x^25 cannot be factorised to give an (x1) bracket, so is not divisible by it and does therefore not fit the sequence. 
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 02032018 18:31
(Original post by irrelevant kid)
Okay, thanks. Is this right?
Just using a term in the sequence to show whether I know a term fits:
2x=(x1)n+2
2x2=(x1)n
2(x1)=(x1)n
So n=2, so fits the sequence.
In the case of 4x^23:
4x^23=(x1)n+2
4x^25=(x1)n
4x^25 cannot be factorised to give an (x1) bracket, so is not divisible by it and does therefore not fit the sequence. 
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 02032018 18:35
(Original post by the bear)
you would need to show a bit more detail of "4x^25 cannot be factorised to give an (x1) bracket", but your idea is very good. 
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 02032018 18:37
(Original post by irrelevant kid)
Okay, thanks a lot! 
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 02032018 18:39
(Original post by the bear)
are you sure it was a GCSE question ? 
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 02032018 18:57
just realized that you can find x quite easily... there is only one suitable value

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 02032018 19:49
(Original post by the bear)
just realized that you can find x quite easily... there is only one suitable value
Posted on the TSR App. Download from Apple or Google PlayLast edited by Y11_Maths; 02032018 at 19:54. Reason: TDA Post Edit 
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 02032018 19:50
(Original post by irrelevant kid)
Okay, thanks. Is this right?
Just using a term in the sequence to show whether I know a term fits:
2x=(x1)n+2
2x2=(x1)n
2(x1)=(x1)n
So n=2, so fits the sequence.
In the case of 4x^23:
4x^23=(x1)n+2
4x^25=(x1)n
4x^25 cannot be factorised to give an (x1) bracket, so is not divisible by it and does therefore not fit the sequence.Posted on the TSR App. Download from Apple or Google Play 
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 02032018 19:52
(Original post by Y11_Maths)
This is incorrect 
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 02032018 19:55
(Original post by the bear)
it is correct, but not the best method.Posted on the TSR App. Download from Apple or Google Play 
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 02032018 19:56
(Original post by Y11_Maths)
You won’t obtain any marks for this method though because it is not what the question is asking for 
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 02032018 20:00
(Original post by the bear)
the question does not specifically ask you to find x.Posted on the TSR App. Download from Apple or Google Play 
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 02032018 20:01
(Original post by Y11_Maths)
It is a 6 mark question so I would edge my bets. I posted how the question is meant to be answered above and I have answered this question before and with all the steps it takes it definitely wants you to find x and solve. 
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 02032018 20:03
(Original post by the bear)
if you still think the other method does not work please explain why.Posted on the TSR App. Download from Apple or Google Play 
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 02032018 20:04
(Original post by irrelevant kid)
Okay, thanks. Is this right?
Just using a term in the sequence to show whether I know a term fits:
2x=(x1)n+2
2x2=(x1)n
2(x1)=(x1)n
So n=2, so fits the sequence.
In the case of 4x^23:
4x^23=(x1)n+2
4x^25=(x1)n
4x^25 cannot be factorised to give an (x1) bracket, so is not divisible by it and does therefore not fit the sequence.
For example consider x1 and x+1. Clearly x+1 doesn't contain an x1 bracket. But this doesn't necessarily mean x1 isn't a factor.
For example, if x=3 then: x1=2, x+1=4. Obviously 2 is a factor of 4.Posted on the TSR App. Download from Apple or Google Play 
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 02032018 20:05
(Original post by pscmaths)
Just because 4x^25 can't be factored to give an x1 bracket, it doesn't mean that x1 doesn't divide 4x^25.
For example consider x1 and x+1. Clearly x+1 doesn't contain an x1 bracket. But this doesn't necessarily mean x1 isn't a factor.
For example, if x=3 then: x1=2, x+1=4. Obviously 2 is a factor of 4.Posted on the TSR App. Download from Apple or Google Play 
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 02032018 21:30
(Original post by Y11_Maths)
This is incorrect
2x  (x+1)
Difference = x1
5x3(x^22)
Difference = 5xx^21
x1=5xx^21
x^24x=0
x(x4)=0
x=4
x+1=5
2x=8
2(2x+3)/6x=11
x^22=14
5x3=17
nth term= 3n+2
4x^23 = 61
3n+2=61
3n=59
59/3 is not an integer so 4x^23 is not a term in the sequence. 
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 02032018 21:33
(Original post by irrelevant kid)
Is this right?
2x  (x+1)
Difference = x1
5x3(x^22)
Difference = 5xx^21
x1=5xx^21
x^24x=0
x(x4)=0
x=4
x+1=5
2x=8
2(2x+3)/6x=11
x^22=14
5x3=17
nth term= 3n+2
4x^23 = 61
3n+2=61
3n=59
59/3 is not an integer so 4x^23 is not a term in the sequence.Posted on the TSR App. Download from Apple or Google Play
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